Mike Holt s Illustrated Guide to ELECTRICAL EXAM PREPARATION

Transcription

1 Mike Holt s Illustrated Guide to LCTRICAL XAM RARATION Theory Calculations Code Based on the 2017 NC

2 CHATR 1 LCTRICAL THORY SSNTIAL FOR JOURNYMAN AND MASTR/CONTRACTOR LICNSING XAMS Unit 1 lectrician s Math and Basic lectrical Formulas Unit 2 lectrical Circuits Unit 3 Understanding Alternating Current Unit 4 Motors and Transformers Note: This chapter is a practice review of theory as it relates to basic electrical formulas and calculations. Most state electrical exams include questions on general electrical theory and you need a solid foundation to pass your exam. If, after working through this chapter, you feel like you need additional in-depth training on theory, then Mike Holt s Theory DVD Training rogram (textbook and DVDs) will give you a comprehensive understanding of this topic. For additional help, visit or call to order your copy. Mike Holt nterprises NC.COD ( ) 7

3 Notes 8 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

4 UNIT 1 LCTRICIAN S MATH AND BASIC LCTRICAL FORMULAS Introduction to Unit 1 lectrician s Math and Basic lectrical Formulas In order to construct a building that will last into the future, a strong foundation is a prerequisite. The foundation is a part of the building that isn t visible in the finished structure, but is essential in erecting one that will have the necessary strength to endure. The math and basic electrical concepts of this unit are very similar to the foundation of a building. The concepts presented here are the essential basics that you must understand, because you ll build upon them as you study electrical circuits and systems. As your studies continue, you ll find that a good foundation in electrical theory and math will help you understand why the NC contains certain provisions. This unit includes math and electrical fundamentals. You ll be amazed at how often your electrical studies return to the basics of this unit. Ohm s law and the electrical formulas related to it are the foundation of all electrical circuits. very student begins at a different level of understanding, and you may find this unit an easy review, or you may find it requires a high level of concentration. In any case, be certain that you fully understand these concepts and are able to successfully complete the questions at the end of this unit before going on. A solid foundation will help in your successful study of the rest of this textbook. art A lectrician s Math Numbers can take different forms: Introduction Whole numbers: 1, 20, 300, 4,000, 5,000 Decimals: 0.75, 0.80, 1.15, 1.25 Fractions: 1/2, 1/4, 5 8, 4 3 ercentages: 80%, 125%, 250%, 500% You ll need to be able to convert these numbers from one form to another and back again, because all of these number forms are part of electrical work and electrical calculations. 1.1 Whole Numbers Whole numbers are exactly what the term implies. These numbers don t contain any fractions, decimals, or percentages. Another name for whole numbers is integers. 1.2 Decimals The decimal method is used to display numbers other than whole numbers, fractions, or percentages such as, 0.80, 1.25, 1.732, and so on. You ll also need to be able to do some basic algebra. Many people are apprehensive of algebra, but as you work through the material here you ll see there s nothing to fear. Mike Holt nterprises NC.COD ( ) 9

5 Unit 1 lectrician s Math and Basic lectrical Formulas 1.3 Fractions A fraction represents part of a whole number. If you use a calculator for adding, subtracting, multiplying, or dividing, you ll need to convert the fraction to a decimal or whole number. To do so, divide the numerator (the top number) by the denominator (the bottom number). }Figure / 4 Fractions 1/ 2 3/ 4 ercentages, Converting to Decimals, xample ercentage Drop % Decimal 32.50% 0 325% % 080% % 125% % 250% 250 Move the decimal point two places to the left. 1 1/ 4 1/ 2 3/ 4 }Figure Multiplier }Figure 1 1 A fraction represents part of a whole, such as 1/ 4, 1/, 2 and 3/. 4 When a number needs to be changed by multiplying it by a percentage, the percentage is called a multiplier. The first step is to convert the percentage to a decimal, then multiply the original number by the decimal value. xamples 1 6 = one divided by six = = two divided by five = = three divided by six = = five divided by four = = seven divided by two = ercentages A percentage is another method used to display a value. One hundred percent (100%) means all of a value, fifty percent (50%) means one-half of a value, and twenty-five percent (25%) means one-fourth of a value. For convenience in multiplying or dividing by a percentage, convert the percentage value to a whole number or decimal, and then use the result for the calculation. When changing a percent value to a decimal or whole number, drop the percentage symbol and move the decimal point two places to the left. }Figure 1 2 xample 1 Question: An overcurrent protection device (circuit breaker or fuse) must be sized no less than 125 percent of the continuous load. If the load is 80A, the overcurrent protection device will have to be sized no smaller than. }Figure 1 3 (a) 75A (b) 80A (c) 100A (d) 125A Using a Multiplier xample Conductors must be sized no less than 125 percent of the continuous load. Convert 125% to a decimal = 1.25 multiplier Multiply 80A x 1.25 = 100A device 80A Continuous Load From Concentrate Jalapeno Flavored Soda }Figure Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

6 lectrician s Math and Basic lectrical Formulas Unit 1 Answer: (c) 100A Step 1: Convert 125 percent to a decimal: 1.25 Step 2: Multiply the value of the 80A load by 1.25 = 100A xample 2 Question: The maximum continuous load on an overcurrent protection device is limited to 80 percent of the device rating. If it s rated 50A, what s the maximum continuous load permitted on the overcurrent protection device? }Figure 1 4 (a) 40A (b) 50A (c) 75A (d) 100A Using a Multiplier xample The circuit supplies a 40A continuous load. Continuous loads are limited to 80% of the overcurrent device rating. Convert 80% to a decimal = 0.80 multiplier 50A device x 0.80 = 40A continuous load 50A Overcurrent Device xample 1 How do you increase the whole number 16 by 25 percent? Step 1: Convert 25 percent to decimal form: 0.25 Step 2: Add one to the decimal value to obtain the multiplier: = 1.25 Step 3: Multiply 16 by the multiplier 1.25: = 20 xample 2 Question: If the feeder demand load for a range is 8 kw and it s required to be increased by 15 percent, the total calculated load will be. }Figure 1 5 (a) 6.80 kw (b) 8 kw (c) 9.20 kw (d) 15 kw Range Rated 8 kw ercent Increase xample To increase 8 kw by 15 percent: Convert 15 percent to a decimal = 0.15 Add a 1 = 1.15 multiplier 8 kw x 1.15 = 9.20 kw calculated load }Figure 1 4 Answer: (a) 40A Step 1: Convert 80 percent to a decimal: 0.80 }Figure 1 5 Step 2: Multiply the value of the 50A device rating by 0.80 = 40A 1.6 ercent Increase Use the following steps to increase a number by a specific percentage: Answer: (c) 9.20 kw Step 1: Convert the percentage increase required to decimal form: 15 percent = 0.15 Step 2: Add one to the decimal: = 1.15 Step 3: Multiply 8 by the multiplier 1.15: 8 kw 1.15 = 9.20 kw Step 1: Convert the percentage to a decimal value. Step 2: Add one to the decimal value to create the multiplier. Step 3: Multiply the original number by the multiplier found in Step 2. Mike Holt nterprises NC.COD ( ) 11

7 Unit 1 lectrician s Math and Basic lectrical Formulas 1.7 Reciprocals To obtain the reciprocal of a number, convert it into a fraction with the number one as the numerator (the top number). It s also possible to calculate the reciprocal of a percentage number. Determine the reciprocal of a percentage number by following these steps: Step 1: Convert the number to a decimal value. Step 2: Divide the value into the number one. xample 1 Question: What s the reciprocal of 80 percent? (a) 0.80 (b) 100% (c) 125% (d) 150% Answer: (c) 125% Step 1: Convert 80 percent into a decimal (move the decimal two places to the left): 80 percent = Squaring a Number Squaring a number means multiplying the number by itself = = 100 or 23 2 = = 529 xample 1 Question: What s the power consumed in watts by a 12 AWG conductor that s 200 ft long, and has a total resistance (R) of 0.40 ohms, if the current (I) in the circuit conductors is 16A? (Answers are rounded to the nearest 50). }Figure 1 6 ower() = I 2 R (a) 50W (b) 100W (c) 150W (d) 200W Squaring a Number, xample 16 Amps Step 2: Divide 0.80 into the number one: 1/0.80 = 1.25 or 125 percent xample 2 Question: What s the reciprocal of 125 percent? (a) 75% (b) 0.80 (c) 100% (d) 125% Answer: (b) 0.80 Step 1: Convert 125 percent into a decimal: 125 percent = 1.25 Step 2: Divide 1.25 into the number one: 1/1.25 = 0.80 or 80 percent }Figure 1 6 Answer: (b) 100W = I 2 R I = 16A R = 0.40 ohms Formula: = I2 x R I (Intensity, Amps) = 16A R (Circuit Resistance) = 0.40 ohms = 16A 2 x 0.40 ohms = ( 16A x 16A) x 0.40 ohms = 256A x 0.40 ohms = W Squaring a number is the value derived when you multiply the number by itself. = 16A ohms = (16A 16A) 0.40 ohms = W Circuit Resistance 0.40 ohms 12 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

8 lectrician s Math and Basic lectrical Formulas Unit 1 xample 2 Question: What s the area in square inches (sq in.) of a trade size 1 raceway with an inside diameter of in.? Area = π r 2 π = 3.14 r = radius (equal to 0.50 of the diameter) (a) 0.34 sq in. (b) 0.50 sq in. (c) 0.86 sq in. (d) 1 sq in. Answer: (c) 0.86 sq in. Area = π r 2 Area = 3.14 ( ) 2 Area = Area = 3.14 ( ) Area = Area = 0.86 sq in. xample 4 Question: What s the sq in. area of a 16-in. pizza? }Figure 1 8 (a) 100 sq in. (b) 150 sq in. (c) 200 sq in. (d) 256 sq in. 8-in. Diameter 50 sq in. If you double the diameter of a circle, the area increases by a factor of four. Squaring, xample 16-in. Diameter Area =πxr 2 Area=3.14 x(0.50 x 16) 2 Area=3.14 x8 2 Area=3.14 x(8x8) Area=3.14 x64 Area=200 sq in. xample 3 Question: What s the sq in. area of an 8-in. pizza? }Figure 1 7 (a) 25 sq in. (b) 50 sq in. (c) 64 sq in. (d) 75 sq in. Squaring, xample 8-in. Diameter 16-in. Diameter }Figure 1 8 Answer: (c) 200 sq in. Area = π r 2 Area = 3.14 ( ) 2 Area = Area = 3.14 (8 8) Area = Area = 200 sq in. }Figure 1 7 Answer: (b) 50 sq in. Area = π r 2 Area = 3.14 (0.50 8) 2 Area = Area = 3.14 (4 4) Area = Area = 50 sq in. Area =πxr 2 Area=3.14 x(0.50 x8) 2 Area=3.14 x4 2 Area=3.14 x(4x4) Area=3.14 x16 Area=50 sq in. Author s Comment: As you see in xamples 3 and 4, if you double the diameter of the circle, the area contained within it is increased by a factor of four! By the way, a large pizza is always cheaper per sq in. than a small one. 1.9 arentheses Whenever numbers are in parentheses, complete the mathematical function within the parentheses before proceeding with the rest of the problem. arentheses are used to group steps of a process into the correct order. For instance, adding the sum of 3 and 15 to the product of 4 and 2 equals 26. (3 + 15) + (4 2) = = 26 Mike Holt nterprises NC.COD ( ) 13

9 ON OFF ON OFF Unit 1 lectrician s Math and Basic lectrical Formulas xample Question: What s the current of a 36,000W, 208V, three-phase load? }Figure 1 9 Amperes = Watts /(Volts 1.732) Amperes = (I), Watts = (), Volts = () (a) 50A (b) 100A (c) 150A (d) 360A Math in arentheses xample 3: Following your calculator s instructions, enter the number 3, then press the square root key = ,000: nter the number 1,000, then press the square root key = If your calculator doesn t have a square root key, don t worry about it. For all practical purposes in using this textbook, the only number you need to know the square root of is 3. The square root of 3 equals approximately To add, subtract, multiply, or divide a number by a square root value, determine the decimal value and then perform the math function. M xample 1 Question: 36,000W/(208V 3) is equal to. 208 V1 36,000W Load 3-hase (a) 100A (b) 120A (c) 208A (d) 360A Answer: (a) 100A I = ( x 3) I = 36,000W (208V x 1.732) I = 36,000W = 100A 360 Whenever numbers are in parentheses, complete the mathematical function within them before proceeding with the rest of the problem. Step 1: Determine the decimal value for the 3 = Step 2: Divide 36,000W by (208V 1.732) = 100A }Figure 1 9 Answer: (b) 100A xample 2 Question: The phase voltage of a 120/208V system is equal to 208V/ 3, which is. Step 1: erform the operation inside the parentheses first determine the product of: 208V = 360V Step 2: Divide 36,000W by 360V = 100A (a) 120V (b) 208V (c) 360V (d)480v Answer: (a) 120V Step 1: Determine the decimal value for the 3 = Step 2: Divide 208V by = 120V 1.10 Square Root Deriving the square root of a number ( n) is the opposite of squaring a number. The square root of 36 is a number that, when multiplied by itself, gives the product 36. The 36 equals six, because six, multiplied by itself (which can be written as 6 2 ) equals the number 36. Because it s difficult to do this manually, just use the square root key of your calculator Volume The volume of an enclosure is expressed in cubic inches (cu in.). It s determined by multiplying the length, by the width, by the depth of the enclosure. 14 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

10 lectrician s Math and Basic lectrical Formulas Unit 1 xample Question: What s the volume of a box that has the dimensions of 4 in. 4 in. 1½ in.? }Figure 1 10 (a) 12 cu in. (b) 20 cu in. (c) 24 cu in. (d) 30 cu in. 4 in. 24 cu in. Volume, xample Volume = Cu In. Cu In. = Length x Width x Depth Cu In.=4in. x4in. x1/ 1 2 in. (Convert the fraction to a decimal and add the 1.) 1+(1 divided by 2=1.50 in. Cu In.=4in. x4in. x1.50 Volume = 24 cu in. 4 in. 1/ 1 2 in. Volume in cubic inches can be determined by multiplying the length, by the width, by the depth of the enclosure. xample 1 Question: What s the wattage value for an 8 kw rated range? (a) 8W (b) 800W (c) 4,000W (d) 8,000W Answer: (d) 8,000W To convert a unit value to a k value, divide the number by 1,000 and add the k suffix. xample 2 Question: What s the kw rating of a 300W load? }Figure 1 11 (a) 0.30 kw (b) 30 kw (c) 300 kw (d) 3,000 kw Kilo - Converting Watts to kw, xample }Figure 1 10 Answer: (c) 24 cu in. 1½ = = 24 cu in. 300W Load 300W/1,000 = 0.30 kw Load Author s Comment: The actual volume of a 4 in. square electrical box is less than 24 cu in. because the interior dimensions may be less than the nominal size and corners are often rounded, so the allowable volume is given in the NC, Table (A) Kilo The letter k is used in the electrical trade to abbreviate the metric prefix kilo, which represents a value of 1,000. To convert a number which includes the k prefix into units, multiply the number preceding the k by 1,000. }Figure 1 11 Answer: (a) 0.30 kw kw = Watts/1,000 kw = 300W/1,000 = 0.30 kw Author s Comment: Watts can be converted to kw by dividing the watts by 1,000. The use of the letter k isn t limited to kw. It s also used for kva (1,000 volt-amperes), kcmil (1,000 circular mils) and other units such as kft (1,000 feet). Mike Holt nterprises NC.COD ( ) 15

11 Unit 1 lectrician s Math and Basic lectrical Formulas 1.13 Rounding Off There s no specific rule for rounding off, but rounding to two or three significant digits should be sufficient for most electrical calculations. Numbers below five are rounded down, while numbers five and above are rounded up the fourth number is five or above = rounded up the fourth number is below five = 1.67 rounded down the fourth number is five or above = 22 rounded up the fourth number is below five = 367 rounded down Rounding Answers for Multiple Choice Questions You should round your answers in the same manner as the multiple choice selections given in the question. xample Question: The sum* of 12, 17, 28, and 40 is equal to. (a) 70 (b) 80 (c) 90 (d) 100 Answer: (d) 100 *A sum is the result of adding numbers. The sum of these values equals 97, but that answer isn t listed as one of the choices. The multiple choice selections in this case are rounded off to the closest tens Testing Your Answer for Reasonableness When working with any mathematical calculation, don t just blindly do the calculation and assume it s correct. When you perform a mathematical calculation, you need to know if the answer is greater than or less than the values given in the problem. Always do a reality check to be certain your answer isn t nonsense. ven the best of us make mistakes at times, so always examine your answer to be sure it makes sense! xample Question: The input of a transformer is 300W; the transformer efficiency is 90 percent. What s the transformer output? }Figure 1 12 (a) 270W (b) 300W (c) 333W (d) 500W Input 300 Watts Testing Your Answer for Reasonableness xample Answer: (a) 270W Since the output must be less than the input (300W), you won t have to perform any mathematical calculation because the only multiple choice selection that s less than 300W is (a) 270W. The math to work out the answer is: 300W 0.90 = 270W To check your multiplication, use division: 270W/0.90 = 300W Author s Comment: 90% fficient Output 270 Watts Since output can t be greater than input, you ll know the answer must be less than 300W. There s only one option less than 300W, so no calculation is necessary. }Figure 1 12 Multiple Choice Selections (a) 270W (b) 300W (c) 333W (d) 500W One of the nice things about mathematical equations is that you can usually test to see if your answer is correct. To do this test, substitute your answer back into the equation with which you re working and verify that it indeed equals out correctly. This method of checking your math will become easier once you know more of the formulas and how they relate to each other. 16 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

12 lectrician s Math and Basic lectrical Formulas Unit 1 art B Basic lectrical Circuits and Formulas Introduction Many false notions about the application of Article 250 Grounding and Bonding, and Chapter 3 Wiring Methods and Materials (both in the NC ) arise when people only use Ohm s Law to solve practice problems on paper but lack a real understanding of how that law works and how it should be applied. After completing this unit, you ll have that understanding, and won t be subject to those false notions or the unsafe conditions to which they lead. But we won t stop with Ohm s Law. You re also going to have a high level of proficiency with the power equation. One of the tools for handling the power equation with ease and Ohm s Law is the power wheel. With it, you ll be able to solve all kinds of problems ower Source The power necessary to move electrons out of their orbit around the nucleus of an atom can be produced by chemical, magnetic, photovoltaic, and other means. The two categories of power sources are direct current (dc) and alternating current (ac). Direct Current The polarity and the output voltage from a direct-current power source never change direction. One terminal is negative and the other is positive, relative to each other. Direct-current power is often produced by batteries, direct-current generators, and electronic power supplies. }Figure 1 14 ower Source, Direct Current 1.15 lectrical Circuit A basic electrical circuit consists of the power source, the conductors, and the load. A switch can be placed in series with the circuit conductors to control the operation of the load (turning it on or off). }Figure The lectron Current Flow Theory states that current travels from negative (-) to positive (+). + 0 Constant dc Waveform Load lectrical Circuit rotection Device ower Source Switch Conductor The polarity and the output voltage from adc power source never change direction. One terminal is negative and the other is positive. DC power is often produced by batteries. }Figure 1 14 Conductor Load (Appliance or quipment) An electrical circuit consists of the power source, the conductors, and the load. A protective device and switch can be placed in series with the circuit conductors. }Figure 1 13 Direct current is used for electroplating, street trolley and railway systems, or where a smooth and wide range of speed control is required for a motor-driven application. Direct current is also used for control circuits and electronic instruments. Alternating Current Alternating-current power sources produce a voltage that changes polarity and magnitude. Alternating current is produced by an alternatingcurrent power source such as an alternating-current generator. The major advantage of alternating current over direct current is that voltage can be changed through the use of a transformer. }Figure 1 15 Mike Holt nterprises NC.COD ( ) 17

13 + Magnitude Unit 1 lectrician s Math and Basic lectrical Formulas N S N S N S N S N S (1) (2) (3) (4) (5) ositive Negative ( + ) 0 ( _ ) }Figure 1 15 Author s Comment: Alternating Current (ac) Voltage ac Waveform Alternating current accounts for more than 90 percent of all electric power used throughout the world Conductance Conductance, or conductivity, is the property of a metal that permits current to flow. The best conductors in order of their conductivity are silver, copper, gold, and aluminum. Copper is most often used for electrical applications. }Figure 1 16 Conductance of Conductor Material at 75 o C +Current and 1.18 Circuit Resistance The total resistance of a circuit includes the resistance of the power supply, the circuit wiring, and the load. Appliances such as heaters and toasters use high-resistance conductors to produce the heat needed for the application. Because the resistances of the power source and conductor are so much smaller than that of the load, they re generally ignored in circuit calculations. }Figure 1 17 rotective Device Low Resistance (Can be Less Than 1 Ohm) ower Source has Very Low Resistance Circuit Resistance Wire ft 12 AWG 0.20 ohms Loads Have the Highest Resistance Wire ft 12 AWG 0.20 ohms 1.19 Ohm s Law 190-Ohm Load The total resistance of a circuit includes the power supply, circuit wiring, and the load. The resistance of the power source will be ignored in this textbook. }Figure 1 17 Ohm s Law expresses the relationship between a direct-current circuit s current intensity (I), electromotive force (), and its resistance (R). This is expressed by the formula: I = /R Author s Comment: The symbol Ω represents ohms. The German physicist Georg Simon Ohm ( ) stated that current is directly proportional to voltage, and inversely proportional to resistance. Silver Copper Gold Aluminum Conductance is the property of metal that permits current to flow. The best conductors in order of conductivity are silver, copper, gold, and aluminum. }Figure 1 16 Direct proportion means that changing one factor results in a direct change to another factor in the same direction and by the same magnitude. If the voltage increases 25 percent, the current increases 25 percent in direct proportion (for a given resistance). If the voltage decreases 25 percent, the current decreases 25 percent in direct proportion (for a given resistance). }Figure Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

14 lectrician s Math and Basic lectrical Formulas Unit 1 I = /R Ohm s Law Current is Directly roportional to Voltage. Voltage (R) =10 ohms 12A Current 120V 120V/10 ohms =12A 240V 240V/10 ohms =24A 24A Direct proportion means that changing one factor results in a direct change to another factor in the same direction and by the same magnitude Ohm s Law and Alternating Current Direct Current In a direct-current circuit, the only opposition to current flow is the physical resistance of the material through which it flows. This opposition is called resistance and is measured in ohms. Alternating Current In an alternating-current circuit, there are three factors that oppose current flow: the resistance of the material; the inductive reactance of the circuit; and the capacitive reactance of the circuit. }Figure 1 18 Inverse proportion means that increasing one factor results in a decrease in another factor by the same magnitude, or a decrease in one factor will result in an increase of the same magnitude in another factor. If the resistance increases by 25 percent, the current decreases by 25 percent in inverse proportion (for a given voltage), or if the resistance decreases by 25 percent, the current increases by 25 percent in inverse proportion (for a given voltage). }Figure 1 19 Author s Comment: For now, we ll assume that the effects of inductance and capacitance on the circuit are insignificant and they ll be ignored Ohm s Law Formula Circle Ohm s Law, the relationship between current, voltage, and resistance expressed in the formula, = I R, can be transposed to I = /R or R = /I. In order to use these formulas, two of the values must be known. I = /R Ohm s Law Current is Inversely roportional to Resistance. Resistance () = 120V Current 10 ohms 12A 120V/10 ohms =12A 20 ohms (2x) 30 ohms (3x) 6A 120V/20 ohms =6A 4A (/) 1 2 (/) 1 3 Inverse proportion means that increasing one factor results in a decrease in another factor by the same magnitude, or a decrease in one factor will result in an increase of the same magnitude in another factor. }Figure 1 19 Author s Comment: lace your thumb on the unknown value in }Figure 1 20, and the two remaining variables will show you the correct formula. Current I Voltage R Ohm s Law Circle Ohm s Law Formula Resistance = I x R I = /R R = /I I I I R R R }Figure 1 20 Mike Holt nterprises NC.COD ( ) 19

15 Unit 1 lectrician s Math and Basic lectrical Formulas Current xample Question: 120V supplies a lamp that has a resistance of 192 ohms. What s the current flow in the circuit? }Figure 1 21 (a) 0.50A (b) 0.60A (c) 1.30A (d) 2.50A 120 Volts }Figure 1 21 Answer: (b) 0.60A Step 1: What s the question? What s I? Step 2: What do you know? = 120V, R = 192 ohms Step 3: The formula is I = /R Ohm s Law, Current Intensity xample Known Amps Formula: I = /R Knowns: = 120V, R = 192 ohms 120V I = I = I = 0.625A R 192 ohms? Determine the current of a 120V, 192-ohm circuit. I R 192 ohms Known Determining Voltage Drop With Ohm s Law, xample 50 ft 12 AWG R=0.10 ohms I R 120V Amps Volts 50 ft 12 AWG VD =? R=0.10 ohms Formula: VD = I x R To determine the voltage drop of conductors, use the resistance of conductors. Known: I = 16A (given), R of each conductor = 0.10 ohms VD = I xr VD = 16A x 0.10 ohms = 1.60V VD = 1.60V per conductor Voltage drop of both conductors = 16A x 0.20 ohms = 3.20V Note: Load operates at 120V VD = V }Figure 1 22 Step 1: What s the question? What s VD? Step 2: What do you know about the conductors? I = 16A and R = 0.20 ohms. The NC lists the alternatingcurrent resistance of 1,000 ft of 12 AWG as 2 ohms [Chapter 9, Table 9]. The resistance of 100 ft is equal to 0.20 ohms. }Figure 1 23 Conductor Resistance xample Step 4: The answer is I = 120V/192 ohms Step 5: The answer is I = 0.625A Voltage-Drop xample Question: What s the voltage drop ( VD ) of two 12 AWG conductors (resistance of 0.20 ohms per 100 ft) supplying a 16A load located 50 ft from the power supply? }Figure 1 22 (a) 1.60V (b) 3.20V (c) 16V (d) 32V Answer: (b) 3.20V To determine the resistance of 100 ft of 12 AWG NC Chapter 9, Table 9, 1,000 ft of 12 AWG = 2 ohms 2 ohms/1,000 ft = ohms per ft ohms per ft x 100 ft = 0.20 ohms for 100 ft }Figure 1 23 Step 3: The formula is VD = I R. Step 4: The answer is VD = 16A 0.20 ohms Step 5: The answer is VD = 3.20V 20 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

16 lectrician s Math and Basic lectrical Formulas Unit 1 Resistance xample Question: What s the resistance of two circuit conductors when the conductor voltage drop is 3V and the current flowing in the circuit is 100A? }Figure 1 24 (a) 0.03 ohms (b) 2 ohms (c) 30 ohms (d) 300 ohms Determining Resistance of Conductors xample 1.5 Volts 1.5 Volts R =? Amps 100 I R Author s Comment: lace your thumb on the unknown value in }Figure 1 25 and the two remaining variables will show you the correct formula. ower Current Voltage ower Formula = I x I = / I I I Formula: R = /I Known: of conductors, 1.50V V = 3VD, I = 100A R = I R = 3 VD 100A = 0.03 ohms for both conductors R = ohms x2conductors = 0.03 ohms, both conductors }Figure 1 25 I Circle = /I I }Figure 1 24 Answer: (a) 0.03 ohms Step 1: What s the question? What s R? Step 2: What do you know about the conductors? VD = 3V dropped and I = 100A. Step 3: The formula is R = /I. Step 4: The answer is R = 3V/100A Step 5: The answer is R = 0.03 ohms 1.22 I Formula Circle The I formula circle demonstrates the relationship between power, current, and voltage, and is expressed in the formula = I. This formula can be transposed to I = / or = /I. In order to use these formulas, two of the values must be known. ower Loss xample Question: What s the power loss in watts for two conductors that carry 12A and have a voltage drop of 3.60V? }Figure 1 26 (a) 4.30W (b) 24W (c) 43.20W (d) 432W Determining Conductor ower Loss, xample 1.8 Represents conductor I Volts from source to load. 1.8 Volts }Figure 1 26 =? Amps Formula: = I x Known: I = 12A Known: of conductors = 1.80 VD per conductor = I x = 12A x 1.80 VD = 21.60W per conductor ower is additive: 21.60W x 2 conductors = 43.20W lost OR... = 12A x (1.80 VD VD) = 12A x 3.60 VD = 43.20W lost 12 Mike Holt nterprises NC.COD ( ) 21

17 Unit 1 lectrician s Math and Basic lectrical Formulas Answer: (c) 43.20W Step 1: What s the question? What s? Step 2: What do you know? I = 12A and = 3.60V. Step 3: The formula is = I. Step 4: The answer is = 12A 3.60V. Step 5: The answer is 43.20W. Current xample Question: What s the current flow in amperes through a 7.50 kw heat strip rated 230V when connected to a 230V power supply? }Figure 1 27 (a) 25A (b) 33A (c) 39A (d) 230A 1.23 Formula Wheel The formula wheel is a combination of the Ohm s Law and the I formula wheels. The formulas in the formula wheel can be used for direct-current circuits or alternating-current circuits with unity power factor. }Figure xi ower-watts 2 I xr xr I MF-Volts R () Watts Volts () IxR 7 (R) Ohms Amps (I) 6 I R /R Formula Wheel Resistance-Ohms 2 I Current-Amperes xi 2 I xr xr I R () (R) Watts Ohms Volts Amps () (I) IxR 7 /R 6 I R 2 I ower Formula, Current xample? Amps Known 230 Volts I }Figure 1 28 Author s Comment: The Formula Wheel is a combination of the Ohm s Law and I Circles. }Figure 1 27 Answer: (b) 33A Step 1: What s the question? What s I? Step 2: What do you know? = 7,500W and = 230V. Step 3: The formula is I = /. Step 4: The answer is I = 7,500W/230V. Step 5: The answer is 32.60A. Formula: I = / 7.50 kw Known Known: = 7.50 kw x 1,000 = 7,500W Known: = 230V I = I = 7,500W I = 32.61A 230V Unity power factor is explained in Unit 3. For the purpose of this unit, we ll assume a power factor of 1.00 for all alternatingcurrent circuits Using the Formula Wheel The formula wheel is divided into four sections with three formulas in each. }Figure 1 29 When working the formula wheel, the key to calculating the correct answer is to follow these steps: Step 1: Know what the question is asking for: I,, R, or. Step 2: Determine the knowns: I,, R, or. Step 3: Determine which section of the formula wheel applies: I,, R, or and select the formula from that section based on what you know. Step 4: Work out the calculation. 22 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

18 lectrician s Math and Basic lectrical Formulas Unit 1 () Watts ower-watts MF-Volts Volts () xample Question: The total resistance of two 12 AWG conductors, 75 ft long is 0.30 ohms, and the current through the circuit is 16A. What s the power loss of the conductors? }Figure 1 30 (a) 20W (b) 75W (c) 150W (d) 300W Formula Wheel, xample 75 ft of 12 AWG 0.15 Ohms 0.15 Ohms Answer: (b) 75W ower Loss =? Amps Step 1: What s the question? What s the power loss of the conductors? xi R I 10 2 () (R) I xr Watts Ohms I2 Volts Amps xr () (I) 9 I R 8 IxR 5 /R ft of 12 AWG Formula 10: = I2 x R Known: I = 16A, Known: R of one conductor = 0.15 ohms* *0.15 ohms determined by NC Ch 9, Tbl 9(per ft method) = I2 x R Or = I2 x R = (16A x 16A) x 0.15 ohms = (16A x 16A) x 0.30 ohms = 256A x 0.15 ohms = 256A x 0.30 ohms = 38.40W on one conductor = 76.80W on both conductors = 38.40W x 2 conductors = 76.80W on both conductors }Figure 1 30 Resistance-Ohms (R) Ohms Amps (I) Formula Wheel Current-Amperes 11 xi 10 2 I xr 9 xr I R () Watts Volts () IxR (R) Ohms Amps (I) R /R The Formula Wheel has four main sections covering: for ower (watts) I for Current (intensity, amps) R for Resistance (ohms) for nergy (volts, MF) }Figure I 2 I Step 2: What do you know about the conductors? I = 16A and R = 0.30 ohms. Step 3: What s the formula? = I 2 R Step 4: Calculate the answer: = 16A² 0.30 ohms = 76.80W. The answer is 76.80W ower Losses of Conductors ower in a circuit can be either useful or wasted. Most of the power used by loads such as fluorescent lighting, motors, or stove elements is consumed in useful work. However, the heating of conductors, transformers, and motor windings is wasted work. Wasted work is still energy used; therefore, it must be paid for, so we call wasted work power loss. xample Question: What s the conductor power loss in watts for a 10 AWG conductor that has a voltage drop of 3 percent in a 240V circuit, and carries a current flow of 24A? }Figure 1 31 (a) 17W (b) 173W (c) 350W (d) 450W }Figure Volts Answer: (b) 173W Conductor ower Loss xample Conductors have a 3% voltage drop. =? Amps xi R 10 2 () (R) I xr Watts Ohms Volts Amps xr () (I) 9 I IxR 7 Formula 11: = x I Known: I = 24A, = 240V x 3% = 7.20 VD = x I = 7.20 VD x 24A = W 8 2 I 3 I2 4 R 5 /R 6 Mike Holt nterprises NC.COD ( ) 23

19 Unit 1 lectrician s Math and Basic lectrical Formulas Step 1: What s the problem asking you to find? What s wasted? Step 2: What do you know about the conductors? I = 24A = 240V 3% = 240V 0.03 = 7.20V Step 3: The formula is = I. Step 4: Calculate the answer: = 24A 7.20V = W. The answer is W Cost of ower Since electric bills are based on power consumed in watts, we should understand how to determine the cost of power. xample Question: What does it cost per year (at 8.60 cents per kwh) for the power loss of two 10 AWG circuit conductors that have a total resistance of 0.30 ohms with a current flow of 24A? }Figure 1 32 (a) $1.30 (b) $13 (c) $130 (d) $1,300 Cost of Conductor ower Loss, xample 0.15 Ohms 0.15 Ohms Cost of Conductor ower Loss =? Amps xi R 10 2 I xr 9 xr I 8 IxR () (R) Watts Ohms Volts Amps () (I) 7 I R /R 6 2 I Answer: (c) $130 Step 1: Determine the amount of power consumed: = I 2 R = 24A ohms = W Step 2: Convert the answer in Step 1 to kw: = W/1,000W = kw Step 3: Determine the cost per hour: (0.086 dollars per kwh) kw = dollars per hr Step 4: Determine the dollars per day: dollars per hr (24 hrs per day) = dollars per day Step 5: Determine the dollars per year: dollars per day (365 days per year) = $ per year Author s Comment: That s a lot of money just to heat up two 10 AWG conductors for one circuit. Imagine how much it costs to heat up the conductors for an entire building! Formula 10: = I2 x R Known: I = 24A Cost at 8.60 cents per kw hour Known: R = 0.30 ohms W/1,000 = kw = I2 x R 8.60 cents per k Wh = $0.086 = (24A x 24A) x 0.30 ohms $0.086 x kw = $ per hour = 576A x 0.30 ohms $ x 24 hrs = = W $ per day $ x 365 days = $ per year }Figure Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

20 lectrician s Math and Basic lectrical Formulas Unit 1 Conclusion to Unit 1 lectrician s Math and Basic lectrical Formulas You ve gained skill in working with Ohm s Law and the power equation, and can use the power wheel to solve a wide variety of electrical problems. You also know how to calculate voltage drop and power loss, and can relate the costs in real dollars. As you work through the practice questions, you ll see how well you ve mastered the mathematical concepts and how ready you are to put them to use in electrical formulas. Always remember to check your answer when you re done then you ll know you have the correct answer every time. As useful as these skills are, there s still more to learn. But, your mastery of these basic electrical formulas means you re well prepared. Work through the questions that follow, and go back over the instructional material if you have any difficulty. When you believe you know the material here in Unit 1, you ll be ready to tackle the electrical circuits covered in Unit 2. Mike Holt nterprises NC.COD ( ) 25

21 UNIT 1 UNIT RVIW QUSTIONS The questions in this section are based on the content you have just reviewed. If you struggle with any of the answers, go back and review that section of Unit 1 one more time. ART A LCTRICIAN S MATH 1. The decimal equivalent for the fraction 1/2 is. (a) 0.50 (b) 0.70 (c) 2 (d) 5 2. The approximate decimal equivalent for the fraction 4 18 is. (a) 0.20 (b) 2.50 (c) 3.50 (d) ercentages 3. To change a percent value to a decimal or whole number, drop the percentage sign and move the decimal point two places to the. (a) right (b) left (c) depends (d) none of these 4. The decimal equivalent for 75 percent is. (a) (b) 0.75 (c) 7.50 (d) The decimal equivalent for 225 percent is. (a) (b) 2.25 (c) (d) The decimal equivalent for 300 percent is. (a) 0.03 (b) 0.30 (c) 3 (d) Multiplier 7. The method of increasing a number by another number is done by using a. (a) percentage (b) decimal (c) fraction (d) multiplier 8. An overcurrent device (circuit breaker or fuse) must be sized no less than 125 percent of the continuous load. If the load is 16A, the overcurrent device will have to be sized at no less than. (a) 17A (b) 20A (c) 23A (d) 30A Mike Holt nterprises NC.COD ( ) 27

22 Unit 1 Unit Review Questions 9. The maximum continuous load on an overcurrent device is limited to 80 percent of the device rating. If the overcurrent device is rated 100A, the maximum continuous load is. (a) 72A (b) 80A (c) 90A (d) 125A 1.6 ercent Increase 10. The feeder calculated load for an 8 kw load, increased by 20 percent is. (a) 8 kw (b) 9.60 kw (c) 10 kw (d) 12 kw 1.7 Reciprocals 11. What s the reciprocal of 1.25? (a) 0.80 (b) 1.10 (c) 1.25 (d) A continuous load requires an overcurrent device sized no smaller than 125 percent of the load. What s the maximum continuous load permitted on a 100A overcurrent device? (a) 75A (b) 80A (c) 100A (d) 125A 1.8 Squaring a Number 13. Squaring a number means multiplying the number by itself. 14. What s the power consumed in watts by a 12 AWG conductor that s 100 ft long and has a resistance (R) of 0.20 ohms, when the current (I) in the circuit is 16A? ower = I 2 x R ower = 16A x 0.20 ohms (a) 50W (b) 75W (c) 100W (d) 200W 15. What s the approximate area in square inches of a trade size 2 raceway? Area = π x r 2, π = 3.14, r = radius (½ of diameter) Area = 3.14 x 1 2 (a) 1 sq in. (b) 2 sq in. (c) 3 sq in. (d) 4 sq in. 16. The numeric equivalent of 4 2 is. (a) 2 (b) 8 (c) 16 (d) The numeric equivalent of 12 2 is. (a) 3.46 (b) 24 (c) 144 (d) 1, arentheses 18. What s the maximum distance that two 14 AWG copper conductors can be run if they carry 16A and the maximum allowable voltage drop is 10V? D = (Cmil x VD )/(2 x K x I) D = (4,110 Cmil x 10V)/(2 wires x ohms x 16A) (a) 50 ft (b) 75 ft (c) 100 ft (d) 150 ft 28 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

23 Unit Review Questions Unit What s the current in amperes of an 18 kw, 208V, three-phase load? Current: I = VA/( x 3) Current: I = 18,000W/(208V x 1.732) (a) 25A (b) 50A (c) 100A (d) 150A 1.10 Square Root 20. Deriving the square root of a number is almost the same as squaring a number. 21. What s the approximate square root of 1,000? (a) 3 (b) 32 (c) 100 (d) The square root of 3 is. (a) 1.50 (b) (c) 9 (d) Volume 23. The volume of an enclosure is expressed in, and is calculated by multiplying the length, by the width, by the depth of the enclosure. (a) cubic inches (b) weight (c) inch-pounds (d) none of these 24. What s the volume (in cubic inches) of a 4 x 4 x 1.50 in. box? 1.12 Kilo 25. What s the kw of a 75W load? (a) kw (b) 0.75 kw (c) 7.50 kw (d) 75 kw 1.13 Rounding Off 26. The approximate sum of 2, 7, 8, and 9 is equal to. (a) 20 (b) 25 (c) 30 (d) Testing Your Answer for Reasonableness 27. The output power of a transformer is 100W and the transformer efficiency is 90 percent. What s the transformer input if the output is lower than the input? Input = Output/fficiency Input = 100W/0.90 (a) 90W (b) 100W (c) 110W (d) 125W ART B BASIC LCTRICAL FORMULAS 1.15 lectrical Circuit 28. An electrical circuit consists of the. (a) power source (b) conductors (c) load (a) 20 cu in. (b) 24 cu in. (c) 30 cu in. (d) 33 cu in. Mike Holt nterprises NC.COD ( ) 29

24 Unit 1 Unit Review Questions 29. According to the lectron Flow Theory, electrons leave the terminal of the source, flow through the conductors and load(s), and return to the terminal of the source. (a) positive, negative (b) negative, positive (c) negative, negative (d) positive, positive 1.16 ower Source 30. The polarity and the output voltage from a direct-current power source changes direction. One terminal will be negative and the other will be positive at one moment, then the terminals switch polarity. 31. Direct current is used for electroplating, street trolley and railway systems, or where a smooth and wide range of speed control is required for a motor-driven application. 32. The polarity and the output voltage from an alternating-current power source never change direction. 33. The major advantage of alternating current over direct current is the ease of voltage regulation by the use of a transformer Conductance 34. Conductance is the property that permits current to flow. 35. The best conductors, in order of their conductivity, are gold, silver, copper, and aluminum. 36. Conductance, or conductivity, is the property of metal that permits current to flow. The best conductors in order of their conductivity are. (a) gold, silver, copper, aluminum (b) gold, copper, silver, aluminum (c) silver, gold, copper, aluminum (d) silver, copper, gold, aluminum 1.18 Circuit Resistance 37. The circuit resistance includes the resistance of the. (a) power source (b) conductors (c) load 38. Often the resistances of the power source and conductor are ignored in circuit calculations Ohm s Law 39. The Ohm s Law formula, I = /R, states that current is proportional to the voltage, and proportional to the resistance. (a) indirectly, inversely (b) inversely, directly (c) inversely, indirectly (d) directly, inversely 40. Ohm s Law demonstrates the relationship between circuit. (a) intensity (b) MF (c) resistance 30 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

25 Unit Review Questions Unit Ohm s Law and Alternating Current 41. In a direct-current circuit, the only opposition to current flow is the physical resistance of the material. This opposition is called reactance and is measured in ohms. 42. In an alternating-current circuit, the factor(s) that oppose current flow is(are). (a) resistance (b) inductive reactance (c) capacitive reactance 1.21 Ohm s Law Formula Circle 43. What s the voltage drop of two 12 AWG conductors (0.40 ohms) supplying a 16A load, located 100 ft from the power supply? VD = I x R VD = 16A x 0.40 ohms (a) 1.60V (b) 3.20V (c) 6.40V (d) 12.80V 44. What s the resistance of the circuit conductors when the conductor voltage drop is 7.20V and the current flow is 50A? R = /I R = 7.20V/50A (a) 0.14 ohms (b) 0.30 ohms (c) 3 ohms (d) 14 ohms 1.22 I Formula Circle 45. What s the power loss in watts of a conductor that carries 24A and has a voltage drop of 7.20V? = I x = 24A x 7.20V (a) 175W (b) 350W (c) 700W (d) 2,400W 1.23 Formula Wheel 46. The formulas in the power wheel apply to. (a) direct-current circuits (b) alternating-current circuits with unity power factor (c) direct-current circuits or alternating-current circuits (d) a and b 1.24 Using the Formula Wheel 47. When working any formula, the key to calculating the correct answer is following these four steps: Step 1: Know what the question is asking you to find. Step 2: Determine the knowns of the circuit. Step 3: Select the formula. Step 4: Work out the formula calculation ower Losses of Conductors 48. ower in a circuit can be either useful or wasted. Wasted work is still energy used; therefore it must be paid for, so we call wasted work. (a) resistance (b) inductive reactance (c) capacitive reactance (d) power loss Mike Holt nterprises NC.COD ( ) 31

26 Unit 1 Unit Review Questions 49. The total circuit resistance of two 12 AWG conductors (each 100 ft long) is 0.40 ohms. If the current of the circuit is 16A, what s the power loss of both conductors? = I 2 x R = 16A 2 x 0.40 ohms (a) 75W (b) 100W (c) 300W (d) 600W 1.27 ower Changes with the Square of the Voltage 51. The voltage applied to a resistor dramatically affects the power consumed by that resistor because power is affected in direct proportion to the voltage. 50. What s the conductor power loss for a 120V circuit that has a 3 percent voltage drop and carries a current flow of 12A? = I x = 12A x (120V x 3%) (a) 43W (b) 86W (c) 172W (d) 1,440W 32 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

27 UNIT 1 UNIT CHALLNG QUSTIONS These questions provide additional work on the content you have just learned. These advanced-level problems allow you to really challenge yourself. ART A LCTRICIAN S MATH 1.12 Kilo 1. One kva is equal to. (a) 100 VA (b) 1,000V (c) 1,000W (d) 1,000 VA ART B BASIC LCTRICAL FORMULAS 1.17 Conductance 2. Which of the following is the most conductive? (a) Bakelite. (b) Oil. (c) Air. (d) Salt water Ohm s Law 3. If the contact resistance of a connection increases and the current of the circuit (load) remains the same, then the voltage dropped across the connection will. (a) increase (b) decrease (c) remain the same (d) can t be determined 4. To double the current of a circuit when the voltage remains constant, the R (resistance) must be. (a) doubled (b) reduced by half (c) increased (d) none of these 5. An ohmmeter is being used to test a relay coil. The equipment instructions indicate that the resistance of the coil should be between 30 and 33 ohms. The ohmmeter indicates that the actual resistance is less than 22 ohms. This reading most likely indicates. (a) the coil is okay (b) an open coil (c) a shorted coil (d) a meter problem 1.24 Using the Formula Wheel 6. To calculate the energy consumed in watts by a resistive appliance, you need to know of the circuit. (a) the voltage and current (b) the current and resistance (c) the voltage and resistance (d) any of these pairs of variables Mike Holt nterprises NC.COD ( ) 33

28 Unit 1 Unit Challenge Questions 7. The power consumed by a resistor can be expressed by the formula I 2 x R. If 120V is applied to a 10-ohm resistor, the power consumed will be. = 2 /R (a) 510W (b) 1,050W (c) 1,230W (d) 1,440W 8. ower loss in a circuit because of heat can be determined by the formula. (a) = R x I (b) = I x R (c) = I 2 x R (d) none of these 9. The energy consumed by a 5-ohm resistor is than the energy consumed by a 10-ohm resistor, assuming the current in both cases remains the same. (a) more (b) less 1.24 Using the Formula Wheel 10. What s the power loss of two 10 AWG conductors when the current through the circuit is 16A and the total resistance is 0.18 ohms? = I 2 x R (a) 2.80W (b) 3.80W (c) 46W (d) 55W 1.27 ower Changes with the Square of the Voltage 11. When a 100W, 115V lamp operates at 230V, the lamp will consume approximately when the total resistance is ohms. = 2 /R (a) 150W (b) 300W (c) 400W (d) 450W 12. A 1,500W resistive heater is rated 230V and is connected to a 208V supply. The power consumed by this load at 208V will be approximately when the total resistance is ohms. = 2 /R (a) 1,225W (b) 1,625W (c) 1,750W (d) 1,850W 13. The total resistance of a circuit is ohms. The load has a resistance of 10 ohms and the wire has a resistance of 0.20 ohms. If the current of the circuit is 12A, then the power consumed by the circuit conductors (0.20 ohms) is approximately. = I 2 x R (a) 8W (b) 29W (c) 39W (d) 45W 34 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

29 NC RACTIC QUIZ 1 STRAIGHT ORDR NC ractice Quiz Straight Order [ ] lease use the 2017 Code book to answer the following questions. These questions allow you to test your knowledge of the Code and are sequenced in the order in which their content appears in the NC. They provide you with an opportunity to start testing your Code knowledge, and give you practice answering the types of Code questions you would see on an electrical exam. They ll help you build your core knowledge of the NC. Article 90. Introduction to the National lectrical Code 1. The NC is. (a) intended to be a design manual (b) meant to be used as an instruction guide for untrained persons (c) for the practical safeguarding of persons and property (d) published by the Bureau of Standards 2. Which of the following systems shall be installed and removed in accordance with the NC requirements? (a) Signaling conductors, equipment, and raceways. (b) Communications conductors, equipment, and raceways. (c) lectrical conductors, equipment, and raceways. 3. The NC applies to the installation of. (a) electrical conductors and equipment within or on public and private buildings (b) outside conductors and equipment on the premises (c) optical fiber cables and raceways 4. Utilities may include entities that are designated or recognized by governmental law or regulation by public service/utility commissions. 5. Chapters 1 through 4 of the NC apply. (a) generally to all electrical installations (b) only to special occupancies and conditions (c) only to special equipment and material 6. Chapters 5, 6, and 7 apply to special occupancies, special equipment, or other special conditions and may supplement or modify the requirements in Chapters 1 through Communications wiring such as telephone, antenna, and CATV wiring within a building shall not be required to comply with the installation requirements of Chapters 1 through 7, except where specifically referenced in Chapter 8. Mike Holt nterprises NC.COD ( ) 35

30 Unit 1 NC ractice Quiz Straight Order 8. Factory-installed wiring of listed equipment need not be inspected at the time of installation of the equipment, except to detect alterations or damage. (a) external (b) associated (c) internal CHATR 1. GNRAL RULS Article 100. Definitions 9. Capable of being reached quickly for operation, renewal, or inspections without resorting to portable ladders or the use of tools (other than keys) is known as. (a) accessible (as applied to equipment) (b) accessible (as applied to wiring methods) (c) accessible, readily 10. The NC defines a(n) as a structure that stands alone or that is separated from adjoining structures by fire walls. (a) unit (b) apartment (c) building (d) utility 11. A cable routing assembly is composed of single or connected multiple channels as well as associated fittings, forming a structural system to communications wires and cables, optical fiber and data cables; and Class 2, Class 3, and Type LTC cables; and power-limited fire alarm cables in plenum, riser, and generalpurpose applications. (a) support (b) route (c) protect (d) a and b 12. Communications equipment includes equipment and conductors used for the transmission of. (a) audio (b) video (c) data (d) any of these 13. A communications raceway is an enclosed channel of nonmetallic materials designed for holding communications wires and cables; optical fiber cables; data cables associated with information technology and communications equipment; Class 2, Class 3, and Type LTC cables; and power-limited fire alarm cables in applications. (a) plenum (b) riser (c) general-purpose 14. The selection and installation of overcurrent protective devices so that an overcurrent condition will be localized to restrict outages to the circuit or equipment affected, is called. (a) overcurrent protection (b) interrupting capacity (c) selective coordination (d) overload protection 15. enclosures are constructed so that dust will not enter under specific test conditions. (a) Dust-ignitionproof (b) Dusttight (c) a or b (d) a and b 16. A fixed, stationary, or portable self-contained, electrically operated and/or electrically illuminated utilization equipment with words or symbols designed to convey information or attract attention describes. (a) an electric sign (b) equipment (c) appliances (d) none of these 17. A(n) that performs field evaluations of electrical or other equipment is known as a Field valuation Body (FB). (a) part of an organization (b) organization (c) a or b (d) none of these 36 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

31 NC ractice Quiz Straight Order Unit quipment or materials to which has been attached a(n) of an FB indicating the equipment or materials were evaluated and found to comply with requirements as described in an accompanying field evaluation report is known as field labeled (as applied to evaluated products). (a) symbol (b) label (c) other identifying mark (d) any of these 19. An interactive inverter is an inverter intended for use in parallel with a(n) to supply common loads that may deliver power to the utility. (a) electric utility (b) photovoltaic (V) system (c) battery (d) none of these 20. A is the total components and subsystem that, in combination, converts solar energy into electric energy for connection to a utilization load. (a) photovoltaic system (b) solar array (c) a and b (d) none of these 21. A raceway is an enclosed channel designed expressly for the holding of wires, cables, or busbars, with additional functions as permitted in the Code. 22. A contact device installed at an outlet for the connection of an attachment plug, or for the direct connection of electrical utilization equipment designed to mate with the corresponding contact device, is known as a(n). (a) attachment point (b) tap (c) receptacle (d) wall plug 23. A single receptacle is a single contact device with no other contact device on the same. (a) circuit (b) yoke (c) run (d) equipment 24. A structure is that which is built or constructed, other than equipment. Article 110. Requirements for lectrical Installations 25. Conductors normally used to carry current shall be unless otherwise provided in this Code. (a) bare (b) stranded (c) of copper or aluminum (d) none of these 26. quipment intended to interrupt current at fault levels shall have an interrupting rating at nominal circuit voltage at least equal to the current that is available at the line terminals of the equipment. 27. When protecting equipment against damage from the weather during construction, minimum provisions provided in NFA 5000 Building Construction and Safety Code, the International Building Code (IBC), and the International Residential Code for Oneand Two-Family Dwellings (IRC) can be referenced for additional information. (a) safety (b) flood (c) weather (d) none of these Mike Holt nterprises NC.COD ( ) 37

32 Unit 1 NC ractice Quiz Straight Order 28. Where a tightening torque is indicated as a numeric value on equipment or in installation instructions provided by the manufacturer, a(n) torque tool shall be used to achieve the indicated torque value, unless the equipment manufacturer has provided installation instructions for an alternative method of achieving the required torque. (a) calibrated (b) identified (c) adjustable (d) listed 29. In other than dwelling units, in addition to requirements for field or factory marking of equipment to warn qualified persons of potential electric arc-flash hazards, a permanent label shall be field or factory applied to service equipment rated or more. (a) 600A (b) 1,000A (c) 1,200A (d) 1,600A 30. Service equipment labeling in other than dwelling units shall not be required if an arc-flash label is applied in accordance with industry practice. (a) routine (b) acceptable (c) documented (d) none of these 31. NFA 70, Standard for lectrical Safety in the Workplace, provides guidance, such as determining severity of potential exposure, planning safe work practices, arc-flash labeling, and selecting. (a) personal protective equipment (b) coordinated overcurrent protective devices (c) a and b (d) none of these 32. Acceptable industry practices for equipment labeling are described in NFA 70, Standard for lectrical Safety in the Workplace. This standard provides specific criteria for developing arc-flash labels for equipment that provides, and so forth. (a) nominal system voltage and incident energy levels (b) arc-flash boundaries (c) minimum required levels of personal protective equipment 33. Reconditioned equipment shall be marked with the name, trademark, or other descriptive marking by which the responsible for reconditioning the electrical equipment can be identified, along with the date of the reconditioning. (a) name of the individual (b) approving authority (c) organization (d) listing agency 34. Where caution, warning, or danger signs or labels are required by this Code, the label marking shall warn of the hazards using effective. (a) words (b) colors (c) symbols (d) any combination of words, colors, or symbols 35. at other than dwelling units shall be legibly field marked with the maximum available fault current, include the date the fault-current calculation was performed, and be of sufficient durability to withstand the environment involved. (a) Service equipment (b) Sub panels (c) Motor control centers 36. NFA 70, Standard for lectrical Safety in the Workplace, provides guidance for working space about electrical equipment, such as determining severity of potential exposure, planning safe work practices, arc-flash labeling, and selecting personal protective equipment. 38 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

33 NC ractice Quiz Straight Order Unit Where equipment operating at 1,000 volts, nominal, or less to ground and likely to require examination, adjustment, servicing, or maintenance while energized is required by installation instructions or function to be located in a space with limited access, and where equipment is installed above a lay-in ceiling, there shall be an opening not smaller than. (a) 6 in. x 6 in. (b) 12 in. x 12 in. (c) 22 in. x 22 in. (d) 22 in. x 30 in. 38. Where equipment operating at 1,000 volts, nominal, or less to ground and likely to require examination, adjustment, servicing, or maintenance while energized is required by installation instructions or function to be located in a space with limited access, the width of the working space shall be the width of the equipment enclosure or a minimum of in., whichever is greater. (a) 12 (b) 22 (c) 26 (d) Where equipment operating at 1,000 volts, nominal, or less to ground and likely to require examination, adjustment, servicing, or maintenance while energized is required by installation instructions or function to be located in a space with limited access, all enclosure doors or hinged panels shall be capable of opening a minimum of degrees. (a) 60 (b) 90 (c) 120 (d) Where equipment operating at 1,000 volts, nominal, or less to ground and likely to require examination, adjustment, servicing, or maintenance while energized is required by installation instructions or function to be located in a space with limited access, the space in front of the enclosure shall comply with the depth requirements of Table (A)(1). 41. Illumination shall be provided for all working spaces about service equipment, switchboards, switchgear, panelboards, or motor control centers. (a) over 600V (b) installed indoors (c) rated 1,200A or more (d) using automatic means of control 42. All switchboards, switchgear, panelboards, and motor control centers shall be located in dedicated spaces and protected from damage, and outdoor installations shall be. (a) installed in identified enclosures (b) protected from accidental contact by unauthorized personnel or by vehicular traffic (c) protected from accidental spillage or leakage from piping systems 43. All switchboards, switchgear, panelboards, and motor control centers shall be located in dedicated spaces and protected from damage and the working clearance space for outdoor installations shall include the zone described in. (a) (A) (b) (B) (c) (C) (d) (D) CHATR 2. WIRING AND ROTCTION Article 210. Branch Circuits 44. Article 210 provides the general requirements for. (a) outside branch circuits (b) branch circuits (c) ungrounded conductors (d) feeder calculations 45. In existing installations where a voltage system(s) already exists and a different voltage system is being added, it shall be permissible to mark only the old system voltage. Mike Holt nterprises NC.COD ( ) 39

34 Unit 1 NC ractice Quiz Straight Order 46. Where two or more branch circuits supply devices or equipment on the same yoke or mounting strap, a means to disconnect simultaneously the ungrounded supply conductors shall be provided at the. (a) point where the branch circuits originate (b) location of the device or equipment (c) point where the feeder originates (d) none of these 47. The GFCI protection required by 210.8(A), (B), (C), (D), and () shall be. (a) the circuit breaker type only (b) accessible (c) readily accessible (d) concealed 48. Requirements for branch-circuit GFCI protection for personnel are contained in and requirements for are contained in 422.5(A). 49. For the application of GFCI protection for personnel, when determining distance from receptacles for sinks [210.8(A)(7) and 210.8(B)(5)] and bathtubs or shower stalls [210.8(A)(9)], the distance shall be measured as the path the cord of an appliance connected to the receptacle would follow without piercing a floor, wall, ceiling, or fixed barrier, or passing through a door, doorway, or window. (a) longest (b) shortest (c) most direct (d) none of these 50. GFCI protection shall be provided for all 15A and 20A, 125V, singlephase receptacles in dwelling unit unfinished portions or areas of basements not intended as habitable rooms. (a) feeders (b) appliances (c) motors 40 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

35 NC RACTIC QUIZ 1 RANDOM ORDR NC ractice Quiz Random Order [ ] lease use the 2017 Code book to answer the following questions. To further improve your ability to use your Code book in a testing environment, the following quiz provides 50 NC questions in random order. This randomization allows you to expand your ability to find items in the NC and improve your knowledge of its organization and design. As you progress through this textbook, you ll find additional articles in subsequent randomized quizzes until you ve been exposed to questions from the entire Code book. If you haven t had in-depth Code training and find any of these questions challenging, you may want to explore our Understanding the National lectrical Code products which will provide detailed instruction of the NC and give you true confidence in this part of your job and on your exam. 1. ach disconnecting means shall be legibly marked to indicate its purpose unless located and arranged so. (a) that it can be locked out and tagged (b) it is not readily accessible (c) the purpose is evident (d) that it operates at less than 300 volts-to-ground 2. An optical fiber cable is a factory assembly or field assembly of one or more optical fibers having a(n) covering. (a) conductive (b) nonconductive (c) overall (d) metallic 3. The term Luminaire means a single individual lampholder by itself. 4. The highest current at rated voltage that a device is identified to interrupt under standard test conditions is the. (a) interrupting rating (b) manufacturer s rating (c) interrupting capacity (d) withstand rating 5. All 15A and 20A, 125V, single-phase receptacles installed in crawl spaces at or below grade level of dwelling units shall have GFCI protection. 6. Continuous duty is defined as. (a) when the load is expected to continue for five hours or more (b) operation at a substantially constant load for an indefinitely long time (c) operation at loads and for intervals of time, both of which may be subject to wide variations (d) operation at which the load may be subject to maximum current for six hours or more Mike Holt nterprises NC.COD ( ) 41

36 Unit 1 NC ractice Quiz Random Order 7. Multiwire branch circuits shall. (a) supply only line-to-neutral loads (b) not be permitted in dwelling units (c) have their conductors originate from different panelboards (d) none of these 8. When modifications to the electrical installation occur that affect the maximum available fault current at the service, the maximum available fault current shall be verified or as necessary to ensure the service equipment ratings are sufficient for the maximum available fault current at the line terminals of the equipment. (a) recalculated (b) increased (c) decreased (d) adjusted 9. On a 4-wire, delta-connected system where the midpoint of one phase winding is grounded, only the conductor or busbar having the higher phase voltage-to-ground shall be durably and permanently marked by an outer finish that is in color. (a) black (b) red (c) blue (d) orange 10. A panel, including buses and automatic overcurrent devices, designed to be placed in a cabinet or cutout box and accessible only from the front is known as a. (a) switchboard (b) disconnect (c) panelboard (d) switch 11. A kitchen is defined as an area with a sink and provisions for food preparation and cooking. (a) listed (b) labeled (c) temporary (d) permanent 12. Constructed, protected, or treated so as to prevent rain from interfering with the successful operation of the apparatus under specified test conditions defines the term. (a) raintight (b) waterproof (c) weathertight (d) rainproof 13. As applied to wiring methods, on or attached to the surface, or behind access panels designed to allow access is known as. (a) open (b) uncovered (c) exposed (d) bare 14. All single-phase receptacles rated 150V to ground or less, 50A or less and three-phase receptacles rated 150V to ground or less, 100A or less installed indoors, in other than dwelling units, in wet locations shall be GFCI protected. 15. Grounded conductors AWG or larger can be identified by distinctive white or gray markings at their terminations. (a) 10 (b) 8 (c) 6 (d) Installations of communications equipment that are under the exclusive control of communications utilities, and located outdoors or in building spaces used exclusively for such installations covered by the NC. (a) are (b) are sometimes (c) are not (d) may be 42 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

37 NC ractice Quiz Random Order Unit Where grounded conductors of different systems are installed in the same raceway, cable, or enclosure, the means of identification of the different neutrals shall be documented in a manner that is or be permanently posted where the conductors of different systems originate. (a) available to the AHJ (b) available through the engineer (c) readily available (d) included in the as-built drawings 18. A is an area that includes a basin with a toilet, urinal, tub, shower, bidet, or similar plumbing fixtures. (a) bath area (b) bathroom (c) rest area (d) none of these 19. is a term indicating that there is an intentional delay in the tripping action of the circuit breaker, which decreases as the magnitude of the current increases. (a) Adverse time (b) Inverse time (c) Time delay (d) Timed unit 20. Industry standards are available for application of reconditioned and refurbished equipment. servicing of equipment that remains within a facility should not be considered reconditioning or refurbishing. (a) Normal (b) Incidental (c) mergency (d) none of these 21. A(n) is a point on the wiring system at which current is taken to supply utilization equipment. (a) box (b) receptacle (c) outlet (d) device 22. means acceptable to the authority having jurisdiction. (a) Identified (b) Listed (c) Approved (d) Labeled 23. GFCI protection shall be provided for all 15A and 20A, 125V, singlephase receptacles in dwelling unit kitchens. (a) installed to serve the countertop surfaces (b) within 6 ft from the top inside edge of the bowl of the sink (c) for all receptacles (d) a and b 24. The authority having jurisdiction has the responsibility for. (a) making interpretations of rules (b) deciding upon the approval of equipment and materials (c) waiving specific requirements in the Code and permitting alternate methods and material if safety is maintained 25. In other than dwelling locations, GFCI protection is required for all single-phase receptacles rated 150V to ground or less, 50A or less and three-phase receptacles rated 150V to ground or less, 100A or less in. (a) indoor wet locations (b) locker rooms with associated showering facilities (c) garages, service bays, and similar areas other than vehicle exhibition halls and showrooms 26. A is an accommodation that combines living, sleeping, sanitary, and storage facilities within a compartment. (a) guest room (b) guest suite (c) dwelling unit (d) single-family dwelling Mike Holt nterprises NC.COD ( ) 43

38 Unit 1 NC ractice Quiz Random Order 27. In other than dwelling units, all single-phase receptacles rated 150 volts-to-ground or less, 50A or less and three-phase receptacles rated 150 volts-to-ground or less, 100A or less installed in(on) shall have GFCI protection for personnel. (a) rooftops (b) kitchens (c) bathrooms 28. This Code covers the installation of for public and private premises, including buildings, structures, mobile homes, recreational vehicles, and floating buildings. (a) optical fiber cables (b) electrical equipment (c) raceways 29. The minimum height of working spaces about electrical equipment, switchboards, panelboards, or motor control centers operating at 1,000V, nominal, or less and likely to require examination, adjustment, servicing, or maintenance while energized shall be ft or the height of the equipment, whichever is greater, except for service equipment or panelboards in existing dwelling units that do not exceed 200A. 30. Internal parts of electrical equipment, including, shall not be damaged or contaminated by foreign materials such as paint, plaster, cleaners, abrasives, or corrosive residues. (a) busbars (b) wiring terminals (c) insulators 31. Hazards often occur because of. (a) overloading of wiring systems by methods or usage not in conformity with the NC (b) initial wiring not providing for increases in the use of electricity (c) a and b (d) none of these 32. As used in the NC, equipment includes. (a) fittings (b) appliances (c) machinery 33. A conductor used to connect the system grounded conductor or the equipment to a grounding electrode or to a point on the grounding electrode system is called the conductor. (a) main grounding (b) common main (c) equipment grounding (d) grounding electrode 34. Compliance with either the SI or the inch-pound unit of measurement system shall be permitted. 35. Connection by means of wire-binding screws, studs, and nuts having upturned lugs or the equivalent shall be permitted for AWG or smaller conductors. (a) 12 (b) 10 (c) 8 (d) The NC does not cover electrical installations in ships, watercraft, railway rolling stock, aircraft, or automotive vehicles. 37. GFCI protection shall be provided for all 15A and 20A, 125V, singlephase receptacles installed in a dwelling unit. (a) attic (b) garage (c) laundry area (d) b and c 38. Service conductors originate at the service point and terminate at the service disconnecting means. 44 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC

39 NC ractice Quiz Random Order Unit Working space shall not be used for. (a) storage (b) raceways (c) lighting (d) accessibility 40. Conduit installed underground or encased in concrete slabs that are in direct contact with the earth is considered a location. (a) dry (b) damp (c) wet (d) moist 41. Connected (connecting) to ground or to a conductive body that extends the ground connection is called. (a) equipment grounding (b) bonded (c) grounded 42. A switch constructed so that it can be installed in device boxes or on box covers, or otherwise used in conjunction with wiring systems recognized by the NC is called a switch. (a) transfer (b) motor-circuit (c) general-use snap (d) bypass isolation 43. A conducting object through which a direct connection to earth is established is a. (a) bonding conductor (b) grounding conductor (c) grounding electrode (d) grounded conductor 44. When the Code uses, it means the identified actions are allowed but not required, and they may be options or alternative methods. (a) shall (b) shall not (c) shall be permitted (d) a or b 45. A Class A GFCI protection device is designed to trip when the current to ground is or higher. (a) 4 ma (b) 5 ma (c) 6 ma (d) 7 ma 46. A unit of an electrical system, other than a conductor, that carries or controls electric energy as its principal function is a(n). (a) raceway (b) fitting (c) device (d) enclosure 47. All 15A and 20A, 125V, single-phase receptacles installed in of dwelling units shall have GFCI protection. (a) unfinished attics (b) finished attics (c) unfinished portions or areas of basements not intended as habitable rooms and crawl spaces (d) finished basements 48. A nominal value assigned to a circuit or system for the purpose of conveniently designating its voltage class, such as 120/240V, is called voltage. (a) root-mean-square (b) circuit (c) nominal (d) source 49. A stand-alone system supplies power independently of an electrical production and distribution network. 50. The NC requires that electrical equipment be. (a) installed in a neat and workmanlike manner (b) installed under the supervision of a licensed person (c) completed before being inspected Mike Holt nterprises NC.COD ( ) 45

40 NC CHALLNG QUIZ 1 NC Challenge Quiz [90.1 Chapter 9 ] lease use the 2017 Code book to answer the following questions. These questions are based on content in the entire NC. Initially you may find them extremely challenging, but as you work through each quiz of this type, you ll gain more confidence in how your Code book is organized and how to find the answers you need. 1. The messenger of messenger-supported wiring shall be supported at dead ends and at intermediate locations so as to eliminate on the circuit conductors. (a) static (b) magnetism (c) tension (d) induction 2. All luminaires, lampholders, and any receptacles installed in theater dressing rooms, dressing areas, and makeup areas adjacent to the mirrors and above the dressing table counter(s), shall be controlled by wall switches in the dressing or makeup room(s). 3. A portable electric sign shall not be placed in or within ft from the inside walls of a fountain. (a) 5 (b) 10 (c) 15 (d) Optical fiber riser cables suitable for use in a vertical run in a shaft or from floor to floor include Types. (a) OFN and OFC (b) OFNR and OFCR (c) OFNG and OFCG (d) OFN and OFC 5. The voltage drop on technical power systems for sensitive electronic equipment shall not exceed percent for feeder and branch-circuit conductors combined. (a) 1.50 (b) 2 (c) 2.50 (d) 3 6. Surface nonmetallic raceways and associated fittings shall be supported in accordance with the installation instructions. (a) vendor s (b) supplier s (c) manufacturer s (d) engineer s 46 Mike Holt s Illustrated Guide to lectrical xam reparation, Based on the 2017 NC